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What is the maximum speed that a car can maneuver over a circular turn of radius r = 75.0 m without sliding if the coefficient of static friction is µs = 0.780? (b) What is the maximum speed that a car can maneuver over a circular turn of radius r = 25.0 m without sliding if the coefficient of static friction is µs = 0.120?

User Arcones
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1 Answer

4 votes

Answer:

Step-by-step explanation:

Given

Radius of Track
r_1=75 m

coefficient of Static Friction
\mu _s=0.78

Here centripetal Force is Balanced by Friction Force

thus


(mv^2)/(r)=\mu _sg


(v^2)/(r)=\mu _sg


v=√(\mu _srg)


v=√(0.78* 75* 9.8)


v=23.94 m/s

(b)For
r_2=25 m


\mu _s=0.12


v=√(\mu _sr_2g)


v=√(0.12* 25* 9.8)


v=5.42 m/s

User Philnate
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